Step of Proof: do-apply-p-filter
11,40
postcript
pdf
Inference at
*
I
of proof for Lemma
do-apply-p-filter
:
T
:Type,
P
:(
T
),
f
:(
x
:
T
. Dec(
P
(
x
))),
x
:
T
.
(
can-apply(p-filter(
f
);
x
))
(do-apply(p-filter(
f
);
x
) =
x
)
latex
by (((if (((first_nat 4:n)) = 0) then (Repeat (((D (0)
)
CollapseTHENA (Auto
))
Co
)) else (RepeatFor (first_nat 4:n) (((D (0)
)
CollapseTHENA (Auto
))
)))
)
CollapseTHEN (
Co
RepUR ``can-apply do-apply p-filter`` ( 0)
))
latex
Co
1
:
Co1:
1.
T
: Type
Co1:
2.
P
:
T
Co1:
3.
f
:
x
:
T
. Dec(
P
(
x
))
Co1:
4.
x
:
T
Co1:
(
isl(case
f
(
x
) of inl(
p
) => inl
x
| inr(
p
) => inr
p
))
Co1:
(outl(case
f
(
x
) of inl(
p
) => inl
x
| inr(
p
) => inr
p
) =
x
)
Co
.
Definitions
Type
,
,
Dec(
P
)
,
x
(
s
)
,
f
(
a
)
,
t
T
,
x
:
A
.
B
(
x
)
,
P
Q
,
s
=
t
,
x
:
A
B
(
x
)
,
can-apply(
f
;
x
)
,
p-filter(
f
)
,
do-apply(
f
;
x
)
,
b
Lemmas
decidable
wf
origin